Compute the typical de Broglie wavelength of an electron in a metal at `27^(@)C` and compare it with the mean separation between two electrons in a metal which is given to be about `2 xx 10^(-10) m.`

Explanation

`T=27+273=300 K,K_(B)=1.38xx10^(-23)Jmo,^(-1)K^(-1)`
`m=9.1xx10^(-31)kg,h=6.63xx10^(-34)Js`
Distance between two electron `r=2xx10^(-10)m`
`implies` de-Broglie wavelength of electron,
`lambda=(h)/(sqrt(3mk_(B)T))`
`=(6.63xx10^(-34))/(sqrt(3xx9.1xx10^(-31)xx1.38xx10^(-28)xx300))`
`(6.63xx10^(-34))/(sqrt(11302.3xx10^(-54)))`
`=(6.63xx10^(-34))/(106.31xx10^(-27))`
`=0.06236xx10^(-7)`
`~~6.2xx10^(-9)`m
`implies` Average distance between two electron in metal,
`r=2xx10^(-10)`m
`therefore (lambda)/(r)=(6.2xx10^(-9))/(2xx10^(-10))=31`
Hence average distance between two electron in atom is 31 times de-Broglie wavelength of electron.